Given a binary tree, return the preorder traversal of its nodes‘ values.
Example:
Input:[1,null,2,3]1 2 / 3 Output:[1,2,3]
Follow up: Recursive solution is trivial, could you do it iteratively?
Solution：
很简单，使用栈，先存入右节点，再存入左节点，这样就是先弹出左节点了
1 class Solution { 2 public: 3 vector<int> preorderTraversal(TreeNode* root) { 4 if (root == nullptr)return {}; 5 vector<int>res; 6 stack<TreeNode*>s; 7 s.push(root); 8 while (!s.empty()) 9 { 10 TreeNode* p = s.top(); 11 s.pop(); 12 res.push_back(p->val); 13 if (p->right)s.push(p->right); 14 if (p->left)s.push(p->left); 15 } 16 return res; 17 } 18 };
力扣算法题—144Binary Tree Preorder Traversal
原文：https://www.cnblogs.com/zzw1024/p/11774591.html